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[Java] LeetCode32 Longest Valid Parentheses

2015-04-28 10:02 369 查看
Given a string containing just the characters
'('
and
')'
,
find the length of the longest valid (well-formed) parentheses substring.

For
"(()"
, the longest valid parentheses substring is
"()"
,
which has length = 2.

Another example is
")()())"
, where the longest valid parentheses substring
is
"()()"
, which has length = 4.

题意:给定一串包含括号的字符串,求字长有效的括号串。能顺序匹配的。这一题刚开始想好久~~,用stack来做题,很容易求出字符串中所有有效的括号长度。但是如何求有效子串呢?如果我们能找出无效括号的index,将有效的index减去上一个无效的index,那么就是该字符的有效子串。弄清楚这一点求最大的有效字符串也会变得很容易。

public int longestValidParentheses(String s) {
        if(s==null)return 0;
        int len=s.length();
        int i=0;
        Stack<Integer> stack=new Stack<Integer>();
        char ch;
        int res=0;
        while(i<len)
        {
            ch=s.charAt(i);
            if(ch=='(')
            stack.push(i);//我们变换思路,将括号的index入栈
            else
            {
               if(!stack.isEmpty()&&s.charAt(stack.peek())=='(')//如果是’)',且与stack顶括号匹配时,弹出
               {
                   stack.pop();
                   res=Math.max(stack.isEmpty()?i+1:i-stack.peek(),res);//为空的话,证明前面没有无效括号,将i+1;不为空的,前面有无效字符,减去无效字符的index
               }else
               {
                   stack.push(i);
               }
            }
            i++;
        }
       return res; 
    }
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